I'm currently trying to re pick up physics after a few years off.. but I'm not sure on the current question and I have no idea what it actually means.
My mathematics is not great but it reminds me of poles from complex analysis stuff.
If some one could tell me the answer and more importantly the reason why I would be very grateful. even point me in the direction of further reading if you can. $$ x y'' -\frac2{x(x-2)}y'+x^2y = 0 $$ which is true?
$x=0$ is a regular singular point of this equation.
$x=0$ is an irregular singular point of this equation.
$x=0$ is an ordinary point of this equation.
$x=2$ is an irregular singular point of this equation.
Singular: You can not compute the general highest derivative of the equation at this point due to division-by-zero. Here that is the second derivative. Obviously, $x=0$ and $x=2$ are singular.
Regular singularity: You can, after applying some factor, write the equation as $$ (x-c)^2y''(x)+(x-c)p(x)y'(x)+q(x)y(x)=0 $$ where $p$ and $q$ are continuous in $x=c$. It does not play any role if these functions have additional roots in $x=c$. What makes this situation "regular" is that the equation is essentially an Euler-Cauchy equation $$ (x-c)^2y''(x)+(x-c)p(c)y'(x)+q(c)y(x)=0 $$ with some perturbation terms added. And it is known how solve Euler-Cauchy equations, as they are equivalent to linear equations with constant coefficients.
$x=0$: $$ x^2y''(x)-\frac{2}{x-2}y'(x)+x^3y(x)=0 $$ fails this test in the middle term, thus a irregular singularity.
$x=2$: $$ (x-2)^2y''(x)-(x-2)\frac2xy'(x)+x^2(x-2)^2y(x)=0$$ fits the pattern, thus a regular singularity.